Question

Find the equation of the circle in complex form which touches the line $iz+\overline{z}+1+i=0$ and the lines $(2-i)z=(2+i)\overline{z}$ and $(2+i)z+(i-2)\overline{z}-4i=0$ are the normals of the circle.

• A. $|z-(1+\frac{i}{2})|=\frac{3}{\sqrt{2}}$
• B. $|z-(1-\frac{i}{2})|=\frac{3}{2}$
• C. $|z-(1+\frac{i}{2})|=\frac{3}{2\sqrt{2}}$
• D. $|z-(1-\frac{i}{2})|=\frac{3}{2\sqrt{2}}$

Answer 1

Answer: (C)

$|z-(1+\frac{i}{2})|=\frac{3}{2\sqrt{2}}$

The given normals of circle are
$(2-i)z=(2+i)\overline{z}$ ....... (1)
$(2+i)z+(i-2)\overline{z}-4i=0$ ......... (2)
Replace $\overline{z}$ from (2) with the help of (1)
$(2+i)z+\frac{(i-2)(2-i)}{(2+i)}z-4i=0$
$(2+i)z+\frac{(4i-3)(2-i)}{(2+i)(2-i)}z-4i=0$
$\Rightarrow (2+i)z+\frac{(11i-2)}{5}z=4i$
$\Rightarrow (16i+8)z=20i$
$\Rightarrow z=\frac{5i}{4i+2}$
$=\frac{5i(2-4i)}{(2+4i)(2-4i)}=\frac{10i+20}{20}$
Centre $z=(1+\frac{i}{2})$
($\because$ point of intersection of two normals is a centre of the circle)
Tangent of the circle is
$iz+\overline{z}+(1+i)=0$ (given)
Equation of tangent in standard form
$\equiv \frac{iz}{(1+i)}+\frac{\overline{z}}{(1+i)}+1=0$
$\equiv (1+i)z+(1-i)\overline{z}+2=0$
$\therefore$ Radius of circle $=$ perpendicular distance from centre on the tangent
$r=\frac{|\frac{(1+i)(2+i)}{2}+\frac{(1-i)(2-i)}{2}+2|}{2}$
$\therefore \frac{3}{2\sqrt{2}}$
$\therefore$ Equation of the circle is $|z-(1+\frac{i}{2})|=\frac{3}{2\sqrt{2}}$
Ans: C